What is the Maximum Velocity of a Person Falling from a 35 Meter Tower?

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A 69-kilogram person falling from a 35-meter tower will reach a maximum velocity determined by kinematic equations for uniform acceleration. The mass of the person does not affect the final speed, as all objects fall at the same rate under gravity. To calculate the time of fall, the formula x = 0.5 * a * t^2 can be used, where 'a' is the acceleration due to gravity (9.8 m/s²). The final velocity can then be calculated using v = √(2 * a * x). The discussion emphasizes the importance of attempting a solution before seeking help in homework forums.
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Homework Statement



A 69 kilogram person falls from a 35 meter tower head first to the ground

Homework Equations



What is the maximum velicoity he reaches before impact?

The Attempt at a Solution

 
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See that #3? That's where you make an attempt at a solution.

We don't spoon-feed homework here.
 
Hint: Use your kinematic equations for uniform acceleration.
 
Sorry if I am a dumbum. I'm 57 and have forgotten all my equations, not that I was very good at physics. I am doing research for a stuntman who has survived this fall and want to know what speed he reached.
 
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The https://www.physicsforums.com/showthread.php?t=5374" that you agreed to when you signed up require you to attempt a solution:

"NOTE: You MUST show that you have attempted to answer your question in order to receive help."


Furthermore, we are expressly requested not to do work for people here; this is afterall, and homework help forum.
 
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V = Vo + at
X - Xo = Vot + .5at2
v2 = vo2 + 2a(X - Xo)
X - Xo = .5(Vo + V)t

V is final velocity in units of meters per seconds (m/s)
Vo is initial velocity in units of meters per seconds (m/s)
a is acceleration in units of meters per squared seconds (m/s2)
t is time in seconds (s)
X is final displacement in units of meters (m)
Xo is initial displacement in units of meters (m)
 
The mass does not matter, any falling object will have the same speed after falling 35m.

First you need the time of the fall.

x = fall distance = 35m
a = acceleration due to gravity = 9.8 m/s
t = time of fall.

x = .5 a t^2
t = \sqrt { \frac {2x} a }

Given the time of the fall the speed is

v = at

so v = \sqrt {2 a x}

That should give you all the equations you need. can you take it from here?
 
thank you DaveC426913 and Integral. I am new to this board and appreciate your help.
I will not bother you again.
 
phuketman said:
thank you DaveC426913 and Integral. I am new to this board and appreciate your help.
I will not bother you again.

Feel free to bother. That's what we're here for. :wink: But there's rules is all.
 
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